Spectrum of an element (${\sigma }_A(a)$)

The spectrum of $a$ in a unital complex algebra $A$ isFor a nonunital algebra it is defined in the unitization of an algebra.

The algebra resolvent set consists of those scalars for which $\lambda 1-x$ is invertible in the algebra. It is open by the Neumann series, and the inverse-valued map is the resolvent of an element. A closed subalgebra can have a smaller algebra resolvent set because an ambient inverse need not lie in it.

For a closed unital subalgebra $B$ of a unital Banach algebra $A$ and $x\in B$, the smaller-algebra algebra resolvent set is relatively open and closed in the larger one. Openness follows from the Neumann series; closedness follows from continuity of inversion in $A$ and norm closedness of $B$. Every larger-algebra resolvent component is therefore either retained whole or absorbed into the smaller-algebra Banach algebra spectrum. The unbounded component is always retained, since inverses at large modulus are norm-convergent series in $x$.

For a closed subalgebra of a Banach algebra $B\subseteq A$ with common identity, the algebra resolvent set ${\rho }_B(x)$ is clopen in ${\rho }_A(x)$. A point in ${\rho }_A(x)$ has a small connected disc on which membership of ${\rho }_B(x)$ is constant, so it cannot lie in $\partial {\sigma }_B(x)$. Also ${\sigma }_A(x)\subseteq {\sigma }_B(x)$; a point in the interior of the former is in the interior of the latter. Together these observations prove the boundary inclusion, without assuming a spectral-boundary theorem.

The identity $(x+\alpha e)-\lambda e=x-(\lambda -\alpha )e$ proves the formula directly from invertibility. In a nonzero complex unital Banach algebra, the nonemptiness of the Banach-algebra spectrum implies that if $x$ is a quasinilpotent element, then $\rho (x+e)=1$. Consequently every nonscalar element cannot simultaneously be quasinilpotent: adding the identity preserves nonscalarity while translating its spectrum.

An element of a complex unital Banach algebra is quasinilpotent if its spectral radius is zero, equivalently its spectrum of an element is $\{0\}$. A nilpotent element is quasinilpotent, but the converse need not hold. The Volterra convolution algebra on a finite interval gives powers which never vanish, yet have factorially decreasing norms, making the inverse series for $\lambda e-x$ converge for every $\lambda \ne 0$.

In a unital Banach algebra, for $\lambda \ne 0$ the existence of $R=(\lambda 1-ab{)}^{-1}$ gives $(\lambda 1-ba{)}^{-1}={\lambda }^{-1}(1+bRa)$, as verified by multiplying on either side. Interchanging $a,b$ proves the converse. Zero may differ: for the unilateral shift operator $S$ on ${\ell }^2({\mathbb{N}}_0)$, $S^{\ast }S=1$ has spectrum $\{1\}$ and $S{S}^{\ast }=1-P_0$ has spectrum $\{0,1\}$.

Every element of a nonzero complex unital Banach algebra has a nonempty compact spectrum of an element. If the resolvent of an element existed everywhere, composing it with any bounded linear functional would give a bounded entire function tending to zero at infinity. The Liouville theorem and point separation by the Hahn-Banach theorem would force the resolvent to be zero, contradicting the inverse equation. Nonunital algebras are treated by unitization.

The approximate point spectrum consists of the scalars $\lambda$ for which there are unit vectors $x_k$ satisfying $\Vert (T-\lambda I)x_k\Vert \to 0$. Equivalently, $T-\lambda I$ is not bounded below.

This is a part of the spectrum of a bounded operator.

The bilateral shift $(Tx{)}_n=x_{n+1}$ is unitary. It has empty point spectrum, while its approximate point spectrum is the unit circle. For $|\lambda |=1$, normalized vectors with entries ${\lambda }^n$ on an interval of length $N$ fail to satisfy $Tx=\lambda x$ only at the two endpoints, giving residual norm $\sqrt{2/N}$.

It is the two-sided sequence case of a shift operator.

The resolvent is $R(\lambda ,a)=(\lambda 1-a{)}^{-1}$ on the complement of the spectrum of an element. In a Banach algebra it is analytic there.

For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

For two points $z,w$ in the resolvent set of an operator $A$,In particular, the resolvent operator is differentiable and ${\partial }_{z}R(z,A)=-R(z,A{)}^2$.

For a closed operator, the resolvent is operator-valued holomorphic on its resolvent set. On a Hilbert space its norm is subharmonic and therefore cannot have a strict interior maximum.

The $ϵ$-pseudospectrum consists of spectral points together with resolvent points where $\Vert (A-zI{)}^{-1}\Vert >{ϵ}^{-1}$. Equivalently it is described by a small lower norm.

For a spectral subset isolated by a contour $\Gamma$ in the resolvent set,projects onto its spectral subspace. Norm-small perturbations preserve the rank of this projection.

For a bounded operator on a complex Banach space, partition a disconnected spectrum into two nonempty compact pieces. The holomorphic function equal to one near the first piece and zero near the second defines a Riesz projection. The holomorphic spectral mapping theorem makes it neither zero nor identity. Its closed range is a proper nonzero invariant vector subspace. This need not be an orthogonal projection, and the argument does not require a normal operator.

If a closed unital subalgebra $A\subseteq B$ contains $x$, then ${\sigma }_A(x)$ is obtained from ${\sigma }_B(x)$ by adjoining some bounded connected components of its complement. Membership of $(x-\lambda 1{)}^{-1}$ in $A$ is constant on every connected component of the resolvent set, and it always holds on the unbounded component.

For a closed unital subalgebra $A\subseteq B$ with common identity, its group of invertible elements of a Banach algebra is relatively open in $A\cap G(B)$ by the Neumann series. It is relatively closed because inversion in $B$ is continuous and $A$ is closed. Consequently each connected component of $A\cap G(B)$ either consists entirely of elements invertible in $A$, or entirely of elements whose inverses do not belong to $A$.

For $x$ in a unital Banach algebra and $f$ holomorphic near $\sigma (x)$, the holomorphic functional calculus defineswhere $\Gamma$ winds once around the spectrum. It is a continuous unital algebra homomorphism and satisfies the spectral mapping theorem $\sigma (f(x))=f(\sigma (x))$.

If a compact spectrum of an element avoids zero and zero belongs to its unbounded complementary component, choose a simple arc from zero to infinity avoiding the Banach algebra spectrum. Its complement is a simply connected neighborhood of the Banach algebra spectrum, admitting a holomorphic logarithm. Apply holomorphic functional calculus to that logarithm and the composition rule for holomorphic functional calculus to the exponential to obtain the displayed identity. An invertible matrix over the complex numbers always satisfies the hypothesis because its finite Banach algebra spectrum cannot separate zero from infinity.

If a spectrum of an element splits into disjoint nonempty compact pieces $X,Y$, choose disjoint open neighborhoods and a holomorphic function equal to one on the first and zero on the second. The holomorphic functional calculus sends it to an idempotent with the corresponding zero/one values under every algebra character. It is neither zero nor identity: if one piece were killed, a reciprocal function on the other piece would make an originally spectral coordinate difference invertible.

The exponential series converges absolutely in a unital Banach algebra. Multiplication of absolutely convergent series gives $e^x{e}^{-x}=1$, and gives $e^{x+y}=e^x{e}^y$ when $x,y$ commute. Termwise differentiation gives $e^{tx}=1+tx+O(t^2)$ near zero. In a C-star algebra, if $h=h^{\ast }$ and $t$ is real, taking adjoints termwise gives $(e^{ith}{)}^{\ast }=e^{-ith}$, so this exponential is a Unitary element of a C-star algebra and has norm one. These facts let a norm bound on a functional recover its real values on Hermitian elements.

If the spectrum of an element $x$ avoids $(-\infty ,0]$, its holomorphic functional calculus value for the principal cube root is the unique $y$ with $y^3=x$ and spectrum in $\{z\ne 0:|\arg z|<\pi /3\}$. For uniqueness, any other such $v$ defines a continuous unital algebra homomorphism $f\mapsto {\Theta }_v(f\circ (z\mapsto z^3))$ from the slit-plane holomorphic functions to the algebra. It sends the coordinate to $x$, so continuity and uniqueness of holomorphic functional calculus identify it with ${\Theta }_x$. Applying it to the principal cube root gives $v=y$. This argument does not assume commutativity of the ambient algebra or separation by algebra characters.

If the spectrum of an element $x$ lies in the open right half-plane, holomorphic functional calculus gives the displayed square root with spectrum in that half-plane. It is unique there. Two such roots $y,v$ have character values equal to the unique scalar square root with positive real part. Every character of an algebra is therefore nonzero on $y+v$, so the character description of the spectrum makes $y+v$ invertible. From $(y-v)(y+v)=0$ one obtains $y=v$. Equal character values alone do not prove equality in a possibly nonsemisimple algebra.

The closed unital algebra generated by a bounded operator and all its resolvents is commutative, because the generators commute. It has exactly the original operator spectrum at that operator: all required inverses outside that spectrum were included, and an inverse in the smaller algebra is also an operator inverse. This permits commutative-algebra linear functional calculus without replacing the spectrum by the possibly larger spectrum of a polynomial-generated algebra.

Equip holomorphic functions on an open spectral neighborhood with the compact-open topology. A fixed admissible contour gives a bound on $\Vert {\Theta }_x(f)\Vert$ by a constant times the supremum of $|f|$ on the contour, proving continuity. The Runge theorem makes rational functions with poles outside the open set dense. A unital homomorphism sending the coordinate function to $x$ must send each inverse coordinate difference to the corresponding resolvent, so its rational values and then all its holomorphic values are forced.

The displayed series converges absolutely in a Banach algebra. It is the holomorphic functional calculus value of the analytic branch $z\mapsto \log (1-z)$ on the unit disk. The composition rule for holomorphic functional calculus and $\exp (\log (1-z))=1-z$ give $\exp (\log (1-x))=1-x$. Thus every sufficiently small norm perturbation of the identity is an exponential.

Choose a cycle $\Gamma$ surrounding $\sigma (x)$ and a cycle $\Delta$ surrounding $f(\sigma (x))$, with $f(\Gamma )$ inside the region of winding number one for $\Delta$. The algebra-homomorphism property givesSubstitute this into the contour formula for $g(f(x))$, interchange integrals, and apply the Cauchy integral formula to the inner $\zeta$ integral. This proves composition as an equality of algebra elements, not merely an equality on characters, which need not separate a nonsemisimple algebra.

Let $x_1,x_2$ belong to a commutative unital Banach algebra, have the same Gelfand transform, and have common spectrum $K$. If $f$ is holomorphic near $K$ and $f^{\prime }$ has no zero on $K$, then $f(x_1)=f(x_2)$ implies $x_1=x_2$. Apply the two-variable holomorphic functional calculus to the divided difference of $f$; its Gelfand transform never vanishes, so it is invertible.

The full spectrum of an element is its spectrum together with every bounded component of its complement. Its complement is the unbounded resolvent component, and the maximum modulus principle controls a holomorphic function on each filled hole by its values on the spectral boundary.